Parameterization

Part 01

One Parameter — Curves

A curve in 2D or 3D is described by a single parameter $t$. Instead of the constraint equation $x^2+y^2=1$, we write a vector-valued function:

\mathbf{r}(t) = \begin{pmatrix} x(t) \\ y(t) \\ z(t) \end{pmatrix}, \quad t \in [a, b]

Term Interrogation

Symbol	Physical Meaning
$\mathbf{r}(t)$	Position vector — arrow from origin to the bug's location at "time" $t$
$x(t),\,y(t),\,z(t)$	Three coordinate functions — each gives one component of position
$t$	The parameter — can be actual time, angle, arc length, or just a dummy variable
$[a,\,b]$	The interval — defines where the curve starts and stops
$\mathbf{r}'(t)$	Velocity / tangent vector — direction the bug is moving at time $t$
$\|\mathbf{r}'(t)\|$	Speed — how fast the bug moves; converts $dt$ into arc length $ds$

Interactive: Drag the Parameter

Watch the traced path, the position vector $\mathbf{r}(t)$, and the tangent vector $\mathbf{r}'(t)$ live.

The Derivative: Your Most Important Tool

Differentiate component-by-component — no chain rule tricks needed:

\mathbf{r}(t) = (\cos t,\; \sin t) \;\Longrightarrow\; \mathbf{r}'(t) = (-\sin t,\; \cos t)

Geometric Fact

For the unit circle, $\mathbf{r}'(t) \perp \mathbf{r}(t)$ always — the velocity is perpendicular to the radius. That's the geometry of circular motion, falling straight out of the algebra.

Arc Length Element

$ds = |\mathbf{r}'(t)|\,dt$ — this converts "integrate with respect to parameter" into "integrate with respect to actual distance along the curve." The magnitude $|\mathbf{r}'(t)|$ is a speed correction factor.

Part 02

Two Parameters — Surfaces

Here is the conceptual leap. A curve needed one dial $t$. A surface needs two dials — $u$ and $v$ — because a surface has two independent directions of freedom.

\mathbf{r}(u, v) = \begin{pmatrix} x(u,v) \\ y(u,v) \\ z(u,v) \end{pmatrix}, \quad u \in [a,b],\;\; v \in [c,d]

The $u$-$v$ Intuition

Think of $(u, v)$ as coordinates on a flat map. The function $\mathbf{r}(u,v)$ is the mapping rule that wraps, bends, and stretches that flat map onto the 3D surface. Just like how a flat piece of paper can be rolled into a cylinder.

Parameter Space

A flat rectangle in $(u,v)$

→

$\mathbf{r}(u,v)$ maps to

3D Surface

The bent/curved result

The Partial Derivatives

With two parameters, derivatives split into two directions:

\mathbf{r}_u = \dfrac{\partial \mathbf{r}}{\partial u} = \begin{pmatrix} \partial x/\partial u \\ \partial y/\partial u \\ \partial z/\partial u \end{pmatrix} \qquad \mathbf{r}_v = \dfrac{\partial \mathbf{r}}{\partial v} = \begin{pmatrix} \partial x/\partial v \\ \partial y/\partial v \\ \partial z/\partial v \end{pmatrix}

Symbol	Physical Meaning
$\mathbf{r}_u$	Tangent vector in the $u$-direction — "what direction does the surface go when I increase $u$ slightly?"
$\mathbf{r}_v$	Tangent vector in the $v$-direction — same idea, other direction
$\mathbf{r}_u \times \mathbf{r}_v$	Cross product — perpendicular to the surface (the normal vector). Its magnitude is the area-stretching factor
$\|\mathbf{r}_u \times \mathbf{r}_v\|$	Area element $dS$ — how much a tiny $du\,dv$ patch in parameter space stretches onto the actual surface

Why the Cross Product?

$\mathbf{r}_u$ and $\mathbf{r}_v$ span the tangent plane at each point. Their cross product is perpendicular to that plane (the surface normal) and its magnitude equals the area of the parallelogram they span — which is exactly the local area element $dS$.

Part 03 · Core Intuition

The $u,v$ Map Analogy

The hardest mental jump in surface parameterization is visualizing how a flat 2D parameter rectangle becomes a curved 3D surface. Here are three ways to build that intuition.

Intuition 1 — Roll a Piece of Paper

A cylinder of radius $R$ and height $h$ is just a flat rectangle that's been rolled up. The flat rectangle has coordinates $(u, v)$ where $u$ runs around the circumference (angle) and $v$ runs up the height:

\mathbf{r}(u, v) = \begin{pmatrix} R\cos u \\ R\sin u \\ v \end{pmatrix}, \quad u \in [0, 2\pi],\;\; v \in [0, h]

The left edge of the rectangle ($u=0$) and the right edge ($u=2\pi$) are the same line in 3D — that's the seam where the paper joins. The bottom edge ($v=0$) becomes the bottom circle; the top edge ($v=h$) becomes the top circle.

Intuition 2 — Grid Lines as Families of Curves

When you hold $v$ fixed and vary $u$, you trace a $u$-curve on the surface. When you hold $u$ fixed and vary $v$, you trace a $v$-curve. Together these form the grid you see on the surface — and $\mathbf{r}_u$ is tangent to $u$-curves, $\mathbf{r}_v$ is tangent to $v$-curves.

Intuition 3 — Why Does $|\mathbf{r}_u \times \mathbf{r}_v|$ Appear?

In single-variable parameterization, $|\mathbf{r}'(t)|$ corrected for how fast the bug moved — it was the speed. For surfaces, the analogous correction asks: "How much does a tiny rectangle $du \times dv$ in parameter space actually stretch into on the surface?"

The Stretching Factor Intuition

A tiny patch at $(u_0, v_0)$ in parameter space, with side lengths $du$ and $dv$, maps to a parallelogram in 3D with sides $\mathbf{r}_u\,du$ and $\mathbf{r}_v\,dv$. The area of that parallelogram is $|\mathbf{r}_u \times \mathbf{r}_v|\,du\,dv$. That's where the cross product magnitude comes from — it's the local area magnification factor.

dS = |\mathbf{r}_u \times \mathbf{r}_v|\,du\,dv$ $\iint_S f\,dS = \int_c^d \int_a^b f\bigl(\mathbf{r}(u,v)\bigr)\;|\mathbf{r}_u \times \mathbf{r}_v|\;du\,dv

The Iterated Integral Connection

The outer integral $\int_c^d$ integrates over all $v$ values. The inner integral $\int_a^b$ integrates over all $u$ values. This is an iterated integral — two nested single-variable integrals. The parameterization is what pulled a complicated surface integral down into a plain rectangle in $(u,v)$ space.

Part 04

Worked Examples

Example A — Sphere

A sphere of radius $R$ uses spherical coordinates as parameters. Think of $\phi$ as latitude (angle from north pole) and $\theta$ as longitude:

\mathbf{r}(\phi, \theta) = \begin{pmatrix} R\sin\phi\cos\theta \\ R\sin\phi\sin\theta \\ R\cos\phi \end{pmatrix}, \quad \phi \in [0, \pi],\;\; \theta \in [0, 2\pi]

Symbol	Meaning
$\phi$	Polar angle — $0$ at north pole, $\pi$ at south pole. Varying $\phi$ with $\theta$ fixed traces a longitude line
$\theta$	Azimuthal angle — goes all the way around. Varying $\theta$ with $\phi$ fixed traces a latitude circle
$R\sin\phi$	The radius of the latitude circle at angle $\phi$ — it's zero at the poles and maximum $R$ at the equator

The surface area element is $dS = R^2 \sin\phi\,d\phi\,d\theta$. The $\sin\phi$ factor accounts for the fact that latitude circles shrink near the poles — less area per $d\phi\,d\theta$ there.

Example B — Graph Surface $z = f(x,y)$

The simplest parameterization: use $x$ and $y$ as their own parameters. Let $u = x$, $v = y$:

\mathbf{r}(u, v) = \begin{pmatrix} u \\ v \\ f(u,v) \end{pmatrix}

Then:

\mathbf{r}_u = \begin{pmatrix} 1 \\ 0 \\ f_u \end{pmatrix}, \quad \mathbf{r}_v = \begin{pmatrix} 0 \\ 1 \\ f_v \end{pmatrix}

The cross product gives:

\mathbf{r}_u \times \mathbf{r}_v = \begin{pmatrix} -f_u \\ -f_v \\ 1 \end{pmatrix} \;\Longrightarrow\; |\mathbf{r}_u \times \mathbf{r}_v| = \sqrt{1 + f_u^2 + f_v^2}

Geometric Meaning

$\sqrt{1+f_x^2+f_y^2}$ is the "slope correction" — it's larger when the surface is steeply tilted, because a flat patch in $xy$-space corresponds to a bigger tilted patch on the surface. Flat surface ($f_x = f_y = 0$) gives $\sqrt{1} = 1$, no correction needed.

Example C — Line Segment (the one you'll use most)

Going from $\mathbf{P}$ to $\mathbf{Q}$:

\mathbf{r}(t) = (1-t)\,\mathbf{P} + t\,\mathbf{Q}, \quad t \in [0,1]

Example D — Torus (Two Circles, Two Parameters)

A torus is what you get when you take a circle of radius $r$ and swing it around an axis at distance $R$. One parameter $u$ goes around the big circle; the other $v$ goes around the tube:

\mathbf{r}(u, v) = \begin{pmatrix} (R + r\cos v)\cos u \\ (R + r\cos v)\sin u \\ r\sin v \end{pmatrix}, \quad u,v \in [0, 2\pi]

Symbol	Meaning
$R$	Major radius — distance from the torus center to the center of the tube
$r$	Minor radius — radius of the tube itself
$u$	Goes around the "donut hole" — vary this, hold $v$ fixed, you trace a big circle
$v$	Goes around the tube — vary this, hold $u$ fixed, you trace a small circle on the tube surface
$R + r\cos v$	The effective radius at angle $v$ — it oscillates between $R-r$ (inner equator) and $R+r$ (outer equator)

Part 05

Strategy Playbook

1

Identify the Shape

Line? Circle? Sphere? Graph? Cylinder? Match to a template.

2

Write $\mathbf{r}$

Plug into the template. Check: does the bug start and end at the right place?

3

Compute $\mathbf{r}'$ or $\mathbf{r}_u, \mathbf{r}_v$

Differentiate component-by-component. No skipped steps.

4

Compute the Magnitude

$|\mathbf{r}'|$ for curves, $|\mathbf{r}_u \times \mathbf{r}_v|$ for surfaces. This is $ds$ or $dS$.

5

Set Up the Integral

Replace $ds$ or $dS$, substitute $\mathbf{r}(t)$ into $f$, integrate over $[a,b]$ or $[a,b]\times[c,d]$.

Quick Reference Table

Shape	Parameterization	Domain	$\|\mathbf{r}'\|$ or $\|\mathbf{r}_u \times \mathbf{r}_v\|$
Line segment $P \to Q$	$\mathbf{r}(t) = \mathbf{P} + t(\mathbf{Q}-\mathbf{P})$	$t \in [0,1]$	$\|\mathbf{Q}-\mathbf{P}\|$ (const.)
Circle radius $R$	$\mathbf{r}(t) = (R\cos t, R\sin t)$	$t \in [0,2\pi]$	$R$
Helix radius $R$, pitch $c$	$\mathbf{r}(t) = (R\cos t, R\sin t, ct)$	$t \in [0, 2\pi n]$	$\sqrt{R^2 + c^2}$
Graph $y=f(x)$	$\mathbf{r}(t) = (t,\, f(t))$	$t \in [a,b]$	$\sqrt{1+[f'(t)]^2}$
Cylinder radius $R$	$\mathbf{r}(u,v) = (R\cos u, R\sin u, v)$	$u \in [0,2\pi]$, $v \in [0,h]$	$R$
Sphere radius $R$	$\mathbf{r}(\phi,\theta) = (R\sin\phi\cos\theta, R\sin\phi\sin\theta, R\cos\phi)$	$\phi \in [0,\pi]$, $\theta \in [0,2\pi]$	$R^2\sin\phi$
Graph $z=f(x,y)$	$\mathbf{r}(u,v) = (u,\, v,\, f(u,v))$	$u,v$ over region $D$	$\sqrt{1+f_u^2+f_v^2}$
Torus $R,r$	$\mathbf{r}(u,v) = ((R+r\cos v)\cos u,\, (R+r\cos v)\sin u,\, r\sin v)$	$u,v \in [0,2\pi]$	$r(R+r\cos v)$

The One-Sentence Truth

Parameterization trades a complicated geometric constraint for a simple interval (or rectangle). Then integration is just calculus you already know, applied over that simple region.

Para­meteri­zation

One Parameter — Curves

Term Interrogation

Interactive: Drag the Parameter

The Derivative: Your Most Important Tool

Two Parameters — Surfaces

The Partial Derivatives

The $u,v$ Map Analogy

Intuition 1 — Roll a Piece of Paper

Intuition 2 — Grid Lines as Families of Curves

Intuition 3 — Why Does $|\mathbf{r}_u \times \mathbf{r}_v|$ Appear?

Worked Examples

Example A — Sphere

Example B — Graph Surface $z = f(x,y)$

Example C — Line Segment (the one you'll use most)

Example D — Torus (Two Circles, Two Parameters)

Strategy Playbook

Quick Reference Table